Question: The equation of a circle $C$ is $x^2+y^2+12x-16y+51 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2+12x) + (y^2-16y) = -51$ $(x^2+12x+36) + (y^2-16y+64) = -51 + 36 + 64$ $(x+6)^{2} + (y-8)^{2} = 49 = 7^2$ Thus, $(h, k) = (-6, 8)$ and $r = 7$.